SUPER LINEAR ALGEBRA

Transcription

1 Super Liear - Cover:Layou 7/7/2008 2:32 PM Page

2 SUPER LINEAR ALGEBRA W. B. Vasaha Kadasamy web: hp://ma.iim.ac.i/~wbv Florei Smaradache INFOLEARNQUEST A Arbor 2008

3 This book ca be ordered i a paper boud repri from: Books o Demad ProQues Iformaio & Learig (Uiversiy of Microfilm Ieraioal) 300 N. Zeeb Road P.O. Box 346, A Arbor MI , USA Tel.: (Cusomer Service) hp://wwwlib.umi.com/bod/ Peer reviewers: Professor Diego Lucio Rapopor Deparameo de Ciecias y Tecologia Uiversidad Nacioal de Quilmes Roque Sae Peña 80, Beral, Bueos Aires, Argeia Dr.S.Osma, Meofia Uiversiy, Shebi Elkom, Egyp Prof. Mircea Euge Selariu, Polyech Uiversiy of Timisoara, Romaia. Copyrigh 2008 by IfoLearQues ad auhors Cover Desig ad Layou by Kama Kadasamy May books ca be dowloaded from he followig Digial Library of Sciece: hp:// ISBN-0: ISBN-3: EAN: Sadard Address Number: Pried i he Uied Saes of America

4 CONTENTS Preface 5 Chaper Oe SUPER VECTOR SPACES 7. Supermarices 7.2 Super Vecor Spaces ad heir Properies 27.3 Liear Trasformaio of Super Vecor Spaces 53.4 Super Liear Algebra 8 Chaper Two SUPER INNER PRODUCT SUPERSPACES Super Ier Produc Super Spaces ad heir properies Superbiliear form Applicaios 24

5 Chaper Three SUGGESTED PROBLEMS 237 FURTHER READING 282 INDEX 287 ABOUT THE AUTHORS 293

6 PREFACE I his book, he auhors iroduce he oio of Super liear algebra ad super vecor spaces usig he defiiio of super marices defied by Hors (963). This book expecs he readers o be well-versed i liear algebra. May heorems o super liear algebra ad is properies are proved. Some heorems are lef as exercises for he reader. These ew class of super liear algebras which ca be hough of as a se of liear algebras, followig a sipulaed codiio, will fid applicaios i several fields usig compuers. The auhors feel ha such a paradigm shif is esseial i his compuerized world. Some oher srucures ough o replace liear algebras which are over a ceury old. Super liear algebras ha use super marices ca sore daa o oly i a block bu i muliple blocks so i is ceraiy more powerful ha he usual marices. This book has 3 chapers. Chaper oe iroduces he oio of super vecor spaces ad eumeraes a umber of properies. Chaper wo defies he oio of super liear algebra, super ier produc spaces ad super biliear forms. Several ieresig properies are derived. The mai applicaio of hese ew srucures i Markov chais ad Leoief ecoomic models

7 are also give i his chaper. The fial chaper suggess 6 problems maily o make he reader udersad his ew cocep ad apply hem. The auhors deeply ackowledge he uflichig suppor of Dr.K.Kadasamy, Meea ad Kama. W.B.VASANTHA KANDASAMY FLORENTIN SMARANDACHE

8 Chaper Oe SUPER VECTOR SPACES This chaper has four secios. I secio oe a brief iroducio abou supermarices is give. Secio wo defies he oio of super vecor spaces ad gives heir properies. Liear rasformaio of super vecor is described i he hird secio. Fial secio deals wih liear algebras.. Supermarices Though he sudy of super marices sared i he year 963 by Paul Hors. His book o marix algebra speaks abou super marices of differe ypes ad heir applicaios o social problems. The geeral recagular or square array of umbers such as A = , B = 4 5 6, C = [3,, 0, -, -2] ad D =

9 are kow as marices. We shall call hem as simple marices [7]. By a simple marix we mea a marix each of whose elemes are jus a ordiary umber or a leer ha sads for a umber. I oher words, he elemes of a simple marix are scalars or scalar quaiies. A supermarix o he oher had is oe whose elemes are hemselves marices wih elemes ha ca be eiher scalars or oher marices. I geeral he kid of supermarices we shall deal wih i his book, he marix elemes which have ay scalar for heir elemes. Suppose we have he four marices; 2 4 a = 0, a = a 2 = 5 7 ad a 22 = Oe ca observe he chage i oaio a ij deoes a marix ad o a scalar of a marix ( < i, j < 2). Le a a2 a = a2 a ; 22 we ca wrie ou he marix a i erms of he origial marix elemes i.e., a = Here he elemes are divided verically ad horizoally by hi lies. If he lies were o used he marix a would be read as a simple marix. 8

10 Thus far we have referred o he elemes i a supermarix as marices as elemes. I is perhaps more usual o call he elemes of a supermarix as submarices. We speak of he submarices wihi a supermarix. Now we proceed o o defie he order of a supermarix. The order of a supermarix is defied i he same way as ha of a simple marix. The heigh of a supermarix is he umber of rows of submarices i i. The widh of a supermarix is he umber of colums of submarices i i. All submarices wih i a give row mus have he same umber of rows. Likewise all submarices wih i a give colum mus have he same umber of colums. A diagrammaic represeaio is give by he followig figure. I he firs row of recagles we have oe row of a square for each recagle; i he secod row of recagles we have four rows of squares for each recagle ad i he hird row of recagles we have wo rows of squares for each recagle. Similarly for he firs colum of recagles hree colums of squares for each recagle. For he secod colum of recagles we have wo colum of squares for each recagle, ad for he hird colum of recagles we have five colums of squares for each recagle. Thus we have for his supermarix 3 rows ad 3 colums. Oe hig should ow be clear from he defiiio of a supermarix. The super order of a supermarix ells us ohig abou he simple order of he marix from which i was obaied 9

11 by pariioig. Furhermore, he order of supermarix ells us ohig abou he orders of he submarices wihi ha supermarix. Now we illusrae he umber of rows ad colums of a supermarix. Example..: Le a = a is a supermarix wih wo rows ad wo colums. Now we proceed o o defie he oio of pariioed marices. I is always possible o cosruc a supermarix from ay simple marix ha is o a scalar quaiy. The supermarix ca be cosruced from a simple marix his process of cosrucig supermarix is called he pariioig. A simple marix ca be pariioed by dividig or separaig he marix bewee cerai specified rows, or he procedure may be reversed. The divisio may be made firs bewee rows ad he bewee colums. We illusrae his by a simple example. Example..2: Le A = is a 6 6 simple marix wih real umbers as elemes. 0

12 A = Now le us draw a hi lie bewee he 2 d ad 3 rd colums. This gives us he marix A. Acually A may be regarded as a supermarix wih wo marix elemes formig oe row ad wo colums. Now cosider A 2 = Draw a hi lie bewee he rows 4 ad 5 which gives us he ew marix A 2. A 2 is a supermarix wih wo rows ad oe colum. Now cosider he marix A 3 = , A 3 is ow a secod order supermarix wih wo rows ad wo colums. We ca simply wrie A 3 as

13 where a a a 2 a a =, a 2 = , a 2 = ad a 22 = The elemes ow are he submarices defied as a, a 2, a 2 ad a 22 ad herefore A 3 is i erms of leers. Accordig o he mehods we have illusraed a simple marix ca be pariioed o obai a supermarix i ay way ha happes o sui our purposes. The aural order of a supermarix is usually deermied by he aural order of he correspodig simple marix. Furher more we are o usually cocered wih aural order of he submarices wihi a supermarix. Now we proceed o o recall he oio of symmeric pariio, for more iformaio abou hese coceps please refer [7]. By a symmeric pariioig of a marix we mea ha he rows ad colums are pariioed i exacly he same way. If he marix is pariioed bewee he firs ad secod colum ad bewee he hird ad fourh colum, he o be symmerically pariioig, i mus also be pariioed bewee he firs ad secod rows ad hird ad fourh rows. Accordig o his rule of symmeric pariioig oly square simple marix ca be 2

14 symmerically pariioed. We give a example of a symmerically pariioed marix a s, Example..3: Le a s = Here we see ha he marix has bee pariioed bewee colums oe ad wo ad hree ad four. I has also bee pariioed bewee rows oe ad wo ad rows hree ad four. Now we jus recall from [7] he mehod of symmeric pariioig of a symmeric simple marix. Example..4: Le us ake a fourh order symmeric marix ad pariio i bewee he secod ad hird rows ad also bewee he secod ad hird colums. a = We ca represe his marix as a supermarix wih leer elemes. 4 3 a = 3 6, a = 4 a 2 = ad a 22 = , so ha 3

15 a = a a a 2 a The diagoal elemes of he supermarix a are a ad a 22. We also observe he marices a ad a 22 are also symmeric marices. The o diagoal elemes of his supermarix a are he marices a 2 ad a 2. Clearly a 2 is he raspose of a 2. The simple rule abou he marix eleme of a symmerically pariioed symmeric simple marix are () The diagoal submarices of he supermarix are all symmeric marices. (2) The marix elemes below he diagoal are he rasposes of he correspodig elemes above he diagoal. The forh order supermarix obaied from a symmeric pariioig of a symmeric simple marix a is as follows. a = a a a a a' a a a a' a' a a a' a' a' a How o express ha a symmeric marix has bee symmerically pariioed (i) a ad a are equal. (ii) a ij (i j); a = a ji ad a ji = a ij. Thus he geeral expressio for a symmerically pariioed symmeric marix; ij a = a a 2... a a' 2 a a 2. M M M a' a' 2... a If we wa o idicae a symmerically pariioed simple diagoal marix we would wrie 4

16 D = D D D 0' oly represes he order is reversed or rasformed. We deoe a = a' ij jus he ' meas he raspose. ij D will be referred o as he super diagoal marix. The ideiy marix I = Is I I r s, ad r deoe he umber of rows ad colums of he firs secod ad hird ideiy marices respecively (zeros deoe marices wih zero as all eries). Example..5: We jus illusrae a geeral super diagoal marix d; d = i.e., d = m 0 0 m. 2 A example of a super diagoal marix wih vecor elemes is give, which ca be useful i experimeal desigs. 5

17 Example..6: Le Here he diagoal elemes are oly colum ui vecors. I case of supermarix [7] has defied he oio of parial riagular marix as a supermarix. Example..7: Le u = u is a parial upper riagular supermarix. Example..8: Le L = ;

18 L is parial upper riagular marix pariioed as a supermarix. Thus T = T a where T is he lower riagular submarix, wih T = ad a' = We proceed o o defie he oio of supervecors i.e., Type I colum supervecor. A simple vecor is a vecor each of whose elemes is a scalar. I is ice o see he umber of differe ypes of supervecors give by [7]. Example..9: Le v = This is a ype I i.e., ype oe colum supervecor. v = v v 2 M v where each v i is a colum subvecors of he colum vecor v. 7

19 Type I row supervecor is give by he followig example. Example..0: v = [ ] is a ype I row supervecor. i.e., v' = [v', v' 2,, v' ] where each v' i is a row subvecor; i. Nex we recall he defiiio of ype II supervecors. Type II colum supervecors. DEFINITION..: Le a = a a2... a m a a a2m a a2... am a = [a a m ] a 2 = [a 2 a 2m ] a = [a a m ] a 2 i.e., a = a M a m is defied o be he ype II colum supervecor. Similarly if a a2 a 2 = a, a 2 = a22,, a m = M M a a2 am a2m M am. Hece ow a = [a a 2 a m ] is defied o be he ype II row supervecor. 8

20 Clearly a 2 a = a = [a a 2 a m ] M a m he equaliy of supermarices. Example..: Le A = be a simple marix. Le a ad b he supermarix made from A. where a = a = 2 6, a 2 = , 2 a 2 = ad a 22 = 0 2. i.e., a = a a a 2 a

21 where b = b = = b b, b 2 = b 2 b , 0 b 2 = [2 0 2 ] ad b 22 = []. ad a = b = We see ha he correspodig scalar elemes for marix a ad marix b are ideical. Thus wo supermarices are equal if ad oly if heir correspodig simple forms are equal. Now we give examples of ype III supervecor for more refer [7]. 20

22 Example..2: ad a = = [T' a'] b = = T b are ype III supervecors. Oe ieresig ad commo example of a ype III supervecor is a predicio daa marix havig boh predicor ad crierio aribues. The ex ieresig oio abou supermarix is is raspose. Firs we illusrae his by a example before we give he geeral case. Example..3: Le a = = a a a a 2 a 2 22 a

23 where a = a 2 = , a 2 = , a 22 = 5 6, 0 2 0, a 3 = The raspose of a ad a 32 = a = a' = Le us cosider he rasposes of a, a 2, a 2, a 22, a 3 ad a 32. a' = a' 2 = 2 0 a = a = 6 2 a' 2 = 2 5 a2 =

24 a' 3 = a' 22 = a' 32 = 2 a3 = a a = 0 = 4 5. a' = a a 2 a 3 a2 a22 a. 32 Now we describe he geeral case. Le a = a a L a a a L a M M M a a L a 2 m m 2 m be a m supermarix. The raspose of he supermarix a deoed by a' = a a 2 L a a 2 a 22 a L 2. M M M a m a 2m L a m a' is a m by supermarix obaied by akig he raspose of each eleme i.e., he submarices of a. 23

25 Now we will fid he raspose of a symmerically pariioed symmeric simple marix. Le a be he symmerically pariioed symmeric simple marix. Le a be a m m symmeric supermarix i.e., a = a a L a a a L a M M M a a L a 2 m 2 22 m2 m 2m mm he raspose of he supermarix is give by a' a' = a (a 2 ) L (a m ) a 2 a' 22 (a 2m ) L M M M a m a 2m L a mm The diagoal marix a are symmeric marices so are ualered by rasposiio. Hece a' = a, a' 22 = a 22,, a' mm = a mm. Recall also he raspose of a raspose is he origial marix. Therefore (a' 2 )' = a 2, (a' 3 )' = a 3,, (a' ij )' = a ij. Thus he raspose of supermarix cosruced by symmerically pariioed symmeric simple marix a of a' is give by a' = a a L a a a L a M M M a a L a 2 m m m 2m mm. 24

26 Thus a = a'. Similarly raspose of a symmerically pariioed diagoal marix is simply he origial diagoal supermarix iself; i.e., if D = d d 2 O d D' = d d 2 O d d' = d, d' 2 = d 2 ec. Thus D = D'. Now we see he raspose of a ype I supervecor. Example..4: Le V = The raspose of V deoed by V' or V is V = [ ]. 25

27 If where v = 3, v 2 = 2 v V = v 2 v ad v 3 = 7 5 V' = [v' v' 2 v' 3 ]. Thus if he v v 2 V = M v V' = [v' v' 2 v' ]. Example..5: Le = = [T a ]. The raspose of i.e., ' = = T a. 26

28 The addiio of supermarices may o be always be defied. Example..6: For isace le ad where a = b = a a b b a 2 a 2 22 b 2 b a = 2, a 2 = 7 a 2 = [4 3], a 22 = [6]. b = [2], b 2 = [ 3] b 2 = 5 2 ad b 22 = I is clear boh a ad b are secod order square supermarices bu here we cao add ogeher he correspodig marix elemes of a ad b because he submarices do o have he same order..2 Super Vecor Spaces ad heir properies This secio for he firs ime iroduces sysemaically he oio of super vecor spaces ad aalyze he special properies associaed wih hem. Throughou his book F will deoe a field i geeral. R he field of reals, Q he field of raioals ad Z p he field of iegers modulo p, p a prime. These fields all are real; whereas C will deoe he field of complex umbers. 27

29 We recall X = (x x 2 x 3 x 4 x 5 x 6 ) is a super row vecor where x i F; F a field; i 6. Suppose Y = (y y 2 y 3 y 4 y 5 y 6 ) wih y i F; i 6 we say X ad Y are super vecors of he same ype. Furher if Z = (z z 2 z 3 z 4 z 5 z 6 ) z i F; i 6 he we do say Z o be a super vecor of same ype as X or Y. Furher same ype super vecors X ad Y over he same field are equal if ad oly if x i = y i for i =, 2,, 6. Super vecors of same ype ca be added he resula is oce agai a super vecor of he same ype. The firs impora resul abou he super vecors of same ype is he followig heorem. THEOREM.2.: This collecio of all super vecors S = {X = (x x 2 x r x r+ x i x i+ x + x +2 x ) x i F}; F a field, i. { < 2 < < r < r + < < i < i+< < + < < } of his ype is a abelia group uder compoe wise addiio. Proof: Le ad X = (x x 2 x r x r+ x i x i+ x + x +2 x ) Y = (y y 2 y r y r+ y i y i+ y + y +2 y ) S. X + Y = {(x + y x 2 + y 2 x r + y r x r+ + y r+ x i + y i x i+ + y i+ x + + y + x +2 + y +2 x + y )} is agai a super vecor of he same ype ad is i S as x i + y i F; i. Clearly ( ) S as 0 F. Now if X = (x x 2 x r x r+ x i x i+ x + x +2 x ) S he X = ( x x 2 x r x r+ x i x i+ x + x +2 x ) S wih X +( X) = ( X) + X = ( ) Also X + Y = Y + X. Hece S is a abelia group uder addiio. 28

30 We firs illusrae his siuaio by some simple examples. Example.2.: Le Q be he field of raioals. Le S = {(x x 2 x 3 x 4 x 5 ) x,, x 5 Q}. Clearly S is a abelia group uder compoe wise addiio of super vecors of S. Take ay wo super vecors say X = ( ) ad Y = ( ) i S. We see X + Y = ( ) ad X + Y S. Also ( ) acs as he super row zero vecor which ca also be called as super ideiy or super row zero vecor. Furher if X = ( ) he X = ( ) is he iverse of X ad we see X + ( X) = ( ). Thus S is a abelia group uder compoewise addiio of super vecors. If X' = ( ) is ay super vecor. Clearly X' S, give i example.2. as X' is o he same ype of super vecor, as X' is differe from X = (x x 2 x 3 x 4 x 5 ). Example.2.2: Cosider he se S = {(x x 2 x 3 x 4 x 5 ) x i Q; i 5}. S is a addiive abelia group. We call such groups as marix pariio groups. Every marix pariio group is a group. Bu every group i geeral is o a pariio group we also call he marix pariio group or super marix group or super special group. Example.2.3: Le S = {(x x 2 x 3 ) x i Q; i 3}. S is a group uder compoe wise addiio of row vecors bu S is o a marix pariio group oly a group. Example.2.4: Le P x x x x x x = x3 x9 x0 x x x 4 2 x i Q; i =, 2,, 2}. 29

31 Clearly P is a group uder marix addiio, which we choose o call as pariio marix addiio. P is a pariio abelia group or we call hem as super groups. Now we proceed o o defie super vecor space. DEFINITION.2.: Le V be a abelia super group i.e. a abelia pariioed group uder addiio, F be a field. We call V a super vecor space over F if he followig codiios are saisfied (i) for all v V ad c F, vc ad cv are i V. Furher vc = cv we wrie firs he field eleme as hey are ermed as scalars over which he vecor space is defied. (ii) for all v, v 2 V ad for all c F we have c(v + v 2 ) = cv + cv 2. (iii) also (v + v 2 ) c = v c + v 2 c. (iv) for a, b F ad v V we have (a + b) v = av + bv also v (a + b) = v a + v b. (v) for every v V ad F,.v = v. = v (vi) (c c 2 ) v = c (c 2 v) for all v V ad c, c 2 F. The elemes of V are called super vecors ad elemes of F are called scalars. We shall illusrae his by he followig examples. Example.2.5: Le V = {(x x 2 x 3 x 4 ) x i R; i 4, he field of reals}. V is a abelia super group uder addiio. Q be he field of raioals V is a super vecor space over Q. For if 0 Q ad v = ( 2 5 3) V; 0v = ( ) V. Example.2.6: Le V = {(x x 2 x 3 x 4 ) x i R, he field of reals i 4}. V is a super vecor space over R. We see here is differece bewee he super vecor spaces meioed i he example.2.5 ad here. We ca also have oher examples. 30

32 Example.2.7: Le y V = y2 y, y 2,y3 Q. y 3 Clearly V is a super group uder addiio ad is a abelia super group. Take Q he field of raioals. V is a super vecor space over Q. Take 5 Q, v = 2 i V. 4 5v = V. As i case of vecor space which depeds o he field over which i has o be defied so also are super vecor space. The followig example makes his more explici. Example.2.8: Le V = y y2 y, y 2,y3 Q; he field of raioal}; y 3 V is a abelia super group uder addiio. V is a super vecor space over Q; bu V is o a super vecor space over he field of reals R. For 2 R; 3

33 v = 5 3 V v = 2 = 2 V as 5 2, 2 ad 3 2 Q. So V is o a super vecor space over R. We ca also have V as a super -uple space. Example.2.9: Le V = {F K F } where F is a field. V is a super abelia group uder addiio so V is a super vecor space over F. Example.2.0: Le V = {(Q 3 Q 3 Q 2 ) = {(x x 2 x 3 y y 2 y 3 z z 2 ) x i, y k, z j Q; i 3; k 3; j 2}. V is a super vecor space over Q. Clearly V is o a super vecor space over he field of reals R. Now as we have marices o be vecor spaces likewise we have super marices are super vecor spaces. Example.2.: Le x x x x x x x x x x x x x x x A = xi Q; i 20 x5 x6 x5 x6 x be he collecio of super marices wih eries from Q. A is a super vecor space over Q. 32

34 Example.2.2: Le x = x x x x V xi R; i 0 x3 x4 x8 x9 x0 V is a super vecor space over Q. Example.2.3: Le. x = x x x x V xi R; i 0 x3 x4 x8 x9 x0 V is a super vecor space over R. V is also a super vecor space over Q. However soo we shall be provig ha hese wo super vecor spaces are differe. Example.2.4: Le a a2 a5 a6 a a a a = a a2 a5 a A ai Q; i 6 a9 a0 a3 a 4. V is a super vecor space over Q. However V is o a super vecor space over R. We call he elemes of he super vecor space V o be super vecors ad elemes of F o be jus scalars. DEFINITION.2.2: Le V be a super vecor space over he field F. A super vecor β i V is said be a liear combiaio of super vecors α,, α i V provided here exiss scalars c,, c i F such ha β = c α + + c α = i i i= cα. 33

35 We illusrae his by he followig example. Example.2.5: Le V = {(a a 2 a 3 a 4 a 5 a 6 ) a i Q; i 6}. V is a super vecor space over Q. Cosider β = ( ) a super vecor i V. Le α = ( 2 0 ), α 2 = ( ) ad α 3 = ( ) be 3 super vecors i V. We ca fid a, b, c i Q such ha aα + bα 2 + cα 3 = β. Example.2.6: Le a c A = a, b, c, d Q. b d A is a super vecor space over Q. Le 2 5 β = 8 A. We have for 2 4, A 4 3 such ha for scalars 4, Q we have = = = β. Now we proceed oo defie he oio of super subspace of a super vecor space V over he field F. 34

36 DEFINITION.2.3: Le V be a super vecor space over he field F. A proper subse W of V is said o be super subspace of V if W iself is a super vecor space over F wih he operaios of super vecor addiio ad scalar muliplicaio o V. THEOREM.2.2: A o-empy subse W of V, V a super vecor space over he field F is a super subspace of V if ad oly if for each pair of super vecors α, β i W ad each scalar c i F he super vecor cα + β is agai i W. Proof: Suppose ha W is a o empy subse of V; where V is a super vecor space over he field F. Suppose ha cα + β belogs o W for all super vecors α, β i W ad for all scalars c i F. Sice W is o-empy here is a super vecor p i W ad hece ( )p + p = 0 is i W. Thus if α is ay super vecor i W ad c ay scalar, he super vecor cα = cα + 0 is i W. I paricular, ( ) α = α is i W. Fially if α ad β are i W he α + β =. α + β is i W. Thus W is a super subspace of V. Coversely if W is a super subspace of V, α ad β are i W ad c is a scalar ceraily cα + β is i W. Noe: If V is ay super vecor space; he subse cosisig of he zero super vecor aloe is a super subspace of V called he zero super subspace of V. THEOREM.2.3: Le V be a super vecor space over he field F. The iersecio of ay collecio of super subspaces of V is a super subspace of V. Proof: Le {W α } be he collecio of super subspaces of V ad le W = I be he iersecio. Recall ha W is defied as α W α he se of all elemes belogig o every W α (For if x W = W α I he x belogs o every W α ). Sice each W α is a super subspace each coais he zero super vecor. Thus he zero super vecor is i he iersecio W ad W is o empy. Le α ad β be super vecors i W ad c be ay scalar. By defiiio of W boh α ad β belog o each W α ad because each W α is a 35

37 super subspace, he super vecor cα + β is i every W α. Thus cα + β is agai i W. By he heorem jus proved; W is a super subspace of V. DEFINITION.2.4: Le S be a se of super vecors i a super vecor space V. The super subspace spaed by S is defied o be he iersecio W of all super subspaces of V which coai S. Whe S is a fiie se of super vecors, ha is S = {α,, α } we shall simply call W, he super subspace spaed by he super vecors {α,, α }. THEOREM.2.4: The super subspace spaed by a o empy subse S of a super vecor space V is he se of all liear combiaios of super vecors i S. Proof: Give V is a super vecor space over he field F. W be a super subspace of V spaed by S. The each liear combiaio α = x α + + x α of super vecors α,, α i S is clearly i W. Thus W coais he se L of all liear combiaios of super vecors i S. The se L, o he oher had, coais S ad is o-empy. If α, β belog o L he α is a liear combiaio. α = x α + + x m α m of super vecors α,, α m i S ad β is a liear combiaio. β = y β + + y m β m of super vecors β j i S; j m. For each scalar, m c α+β = (cx ) α + y β m i i j j i= j= x i, y i F; i, j m. Hece cα + β belogs o L. Thus L is a super subspace of V. Now we have proved ha L is a super subspace of V which coais S, ad also ha ay subspace which coais S coais L. I follows ha L is he iersecio of all super 36

38 subspaces coaiig S, i.e. ha L is he super subspace spaed by he se S. Now we proceed oo defie he sum of subses. DEFINITION.2.5: If S,, S K are subses of a super vecor space V, he se of all sums α + + α K of super vecors α i i S i is called he sum of he subses S, S 2,, S K ad is deoed by S + + S K or by K i= S. i If W,, W K are super subspaces of he super vecor space V, he he sum W = W + W W K is easily see o be a super subspace of V which coais each of super subspace W i. i.e. W is he super subspace spaed by he uio of W, W 2,, W K, i K. Example.2.7: Le x x2 x9 x0 x x x x x x = x7 x8 x8 x9 x A xi Q; i 6 x5 x6 x5 x6 x 7 be a super vecor subspace of V over Q. Le W x x x8 x5 x6 3 = 0 x6 x3 x4 x,x 3,x 6, x 8,x 3,x 4,x 8, x 5, x 6 Q} W is clearly a super subspace of V. Le 37

39 W2 = x 5, x6 Q, x x W 2 is a super subspace of V. Take 0 x2 x9 x0 0 x4 x x2 W3 = x 2, x 9, x 4,x 0,x,x2 Q a proper super vecor subspace of V. Clearly V = W + W 2 + W 3 i.e., x x x x x x x x x x x x x x x x = x x x6 x x 4 0 x8 x5 x x2 x9 x x4 x x2 +. x x The super subspace Wi W I j = ;i j; i, j

40 39 Example.2.8: Le V = {(a b c d e f g h) a, b, c d, e, f, g, h Q} be a super vecor space over Q. Le W = {(a b c 0 e ) a, b, c, e Q}, W is a super space of V. Take W 2 = {(0 0 c 0 0 f g h) f, g, h, c Q}; W 2 is a super subspace of V. Clearly V = W + W 2 ad W W 2 = {(0 0 c ) c Q} is a super subspace of V. I fac W W 2 is also a super subspace of boh W ad W 2. Example.2.9: Le a b c d V a,b,c,d,e,f,g R e f g =. V is a super vecor space over Q. Take 0 0 c d W c,d,e,f R e f 0 =, W is a super subspace of V. Le

41 a b 0 W 0 2 = a,b,g R 0 0 g, W 2 is a super subspace of V. I fac V = W + W 2 ad W W W is he super zero subspace of V. Example.2.20: Le = 2 = V x x x x x x = x7 x4 x9 x0 x x2 such ha x i Q; i 2}, be he super vecor space over Q. Le x x W = x, x 2,x 7,x8 Q x7 x

42 be he super subspace of he super vecor space V x6 W2 = x 6,x 9,x 0,x,x2 Q 0 0 x9 x0 x x2 be a super subspace of he super vecor space V. Clearly V W + W 2. Bu he zero super marix of V = W2 W Now we proceed oo defie he oio of basis ad dimesio of a super vecor space V. DEFINITION.2.6: Le V be a super vecor space over he field F. A subse S of V is said o be liearly depede (or simply depede) if here exiss disic super vecors α, α 2,, α i S ad scalars c, c 2,, c i F, o all of which are zero such ha c α + c 2 α c α = 0. A se which is o liearly depede is called liearly idepede. If he se S coais oly a fiiely may vecors α, α 2,, α we some imes say ha α, α 2,, α are depede (or idepede) isead of sayig S is depede (or idepede). Example.2.2: Le V = {(x x 2 x 3 x 4 x 5 x 6 x 7 ) x i Q; i 7} be a super vecor space over Q. Cosider he super vecors α, α 2,, α 8 of V give by α = ( ) α 2 = ( ) α 3 = ( ) α 4 = ( 0 3 2) α 5 = ( ) α 6 = (8 0 2) α 7 = ( ) 4

43 ad α 8 = ( ). Clearly α, α 2,, α 8 forms a liearly depede se of super vecors of V. Example.2.22: Le V = {(x x 2 x 3 x 4 ) x i Q} be a super vecor space over he field Q. Cosider he super vecor ad α = ( 0 0 0), α 2 = (0 0 0), α 3 = (0 0 0) α 4 = (0 0 0 ). Clearly he super vecors α, α 2, α 3, α 4 form a liearly idepede se of V. If we ake he super vecors ( 0 0 0), (2 0 0) ad ( 4 0 0) hey clearly form a liearly depede se of super vecors i V. DEFINITION.2.7: Le V be a super vecor space over he field F. A super basis or simply a basis for V is clearly a depede se of super vecors V which spas he space V. The super space V is fiie dimesioal if i has a fiie basis. Le V = {(x x r x r+ x k x k+ x )} be a super vecor space over a field F; i.e. x i F; i.. Suppose W = {(x x r )} V he we call W a special super subspace of V. W 2 = {(0 0 x r+ x ) x r+,, x F} is agai a special super subspace of V. W 3 = {( x + x k 0 0) x +,, x k F} is agai a special super subspace of V. We ow illusrae hus siuaio by he followig examples. 42

44 Example.2.23: Le V = {(x x 2 x 3 x 4 x 5 x 6 ) x i Q; i 6} be a super vecor space over Q. The special super subspaces of V are W = {(x ) x Q} is a special super subspace of V. W 2 = {(0 x 2 x 3 x 4 0 0) x 2, x 3, x 4 Q} is a special super subspace of V. W 3 = {( x 5 x 6 ) x 5, x 6 Q} is also a special super subspace of V. W 4 = {(x x2x3x 4 00)} is a special super subspace of V. W 5 = {(x x5 x 6) xx5x6 Q} is a special super subspace of V ad W 6 = {(0 x2 x3 x 4 x5 x 6) x 2,x 3,x 4,x 5,x6 Q} is a special super subspace of V. Thus V has oly 6 special super subspaces. However if P = {(0 x2 0 x 4 0 0) x 2, x4 Q} is oly a super subspace of V ad o a special super subspace of V. Likewise T = {(x 0 x30 x50 x,x 3,x5 Q} is oly a super subspace of V ad o a special super subspace of V. Example.2.24: Le x x6 x x7 x8 x2 x7 x2 x4 x20 V = x3 x8 x3 x2 x 22 xi Q; i 26 x4 x9 x4 x23 x 24 x5 x0 x5 x25 x 26 43

45 be a super vecor space over Q. The special super subspaces of V are as follows. x x W = x x,x 2,x3 Q is a special super subspace of V W2 = x 4,x5 Q x x is a special super subspace of V. 0 x6 x x7 x2 0 0 W3 = 0 x8 x3 0 0 x 6,x,x 7,x 8,x 2,x3 Q is a special super subspace of V W4 = x 9,x 0,x4 adx5 Q 0 x9 x x0 x

46 is a special super subspace of V x7 x x9 x20 W5= x2 x 22 x 7,x 8,x 9,x 20,x2adx22 Q is a special super subspace of V W6 = x 23,x 24,x25 adx26 Q x23 x x25 x 26 is a special super subspace of V. x x W7 = x xo x5 Q x x is also a special super subspace of V. x x6 x 0 0 x2 x7 x2 0 0 W8= x3 x8 x3 0 0 x,x 2,x 3,x 6,x,x 7,x 8,x2 adx3 Q

47 is also a special super subspace of V ad so o, 0 x6 x x7 x8 0 x7 x2 x9 x20 W = 0 x8 x3 x2 x 22 xi Q; 4 i 26 x4 x9 x4 x23 x 24 x5 x0 x5 x25 x 26 is also a special super subspace of V. Now we have see he defiiio ad examples of special super subspace of a super vecor space V. We ow proceed oo defie he sadard basis or super sadard basis of V. 2 Le F be a field V = (F F K F ) be a super vecor space over F. The super vecors, K,,, K, give by + = ( 0 K0 0K0 0 K 00K0) = (0K0 0 K 0 K 0K0) 2 = (0 K 0K0 0 K 0K0) = (0 K0 0K0 K 0K0) + = (0 K0 0K 0 K 0K0) 2 = (0 K0 0K0 K 0 K0) M M M forms a liearly idepede se ad i spas V; so hese super vecors form a basis of V kow as he super sadard basis of V. We will illusrae his by he followig example. Example.2.25: Le V = {(x x 2 x 3 x 4 x 5 ) x i Q; i 5} be a super vecor space over Q. The sadard basis of V is give by 46

48 ad = ( ), 2 = ( ), 3 = ( ), 4 = ( ), 5 = ( ), Example.2.26: Le V = {(x x 2 x 3 x 4 x 5 x 6 x 7 x 8 ) x i Q; i 8} be a super vecor space over Q. The sadard basis for V is give by = ( ), 2 = ( ), 3 = ( ), 4 = ( ), 5 = ( ), 6 = ( ), 7 = ( ), ad 8 = ( ), Clearly i ca be checked by he reader, 2,, 8 forms a super sadard basis of V. Example.2.27: Le x x5 x6 x2 x7 x8 V = xi Q; i 2 x3 x9 x 0 x4 x x 2 be a super vecor space over Q. The sadard basis for V is ; 47

49 =, 2 =, 3 =, =, 5 =, 6 =, = , =, 9 =, = ad = =, 2 The reader is expeced o verify ha, 2,, 2 forms a super sadard basis of V. Now we are goig o give a special oaio for he super row vecors which forms a super vecor space ad he super marices which also form a super vecor space. Le X = (x x x + x k x r+ x ) be a super row vecor wih eries from Q. Defie X = (A A 2 A m ) where each A i is a row vecor A correspods o he row vecors (x x ), he se of row vecors (x + x k ) o A 2 ad so o. Clearly m. Likewise a super marix is also give a special represeaio. 48

50 Suppose x x x x x x x x x x A A A3 A 4 x0 x x2 x22 x 23 A = x x x x x = x x x x x where A is a 3 3 marix give by x x2 x3 x6 x7 A = x4 x5 x 6, A2 = x5 x9 x x x x x is a 3 2 recagular marix A 3 x x x 0 2 = x5 x4 x5 is agai a recagular 2 3 marix wih eries from Q ad A 4 x x = x24 x25 is agai a 2 2 square marix. We see he compoes of a super row vecor are row vecors where as he compoes of a super marix are jus marices. Now we proceed oo prove he followig heorem. THEOREM.2.5: Le V be a super vecor space which is spaed by a fiie se of super vecors β,, β m. The ay idepede se of super vecors i V is fiie ad coais o more ha m elemes. 49

51 Proof: Give V is a super vecor space. To prove he heorem i suffices o show ha every subse S of V which coais more ha m super vecors is liearly depede. Le S be such a se. I S here are disic super vecors α,, α where > m. Sice β, β 2,, β m spa V heir exiss scalars A ij i F such ha m α j = Aijβi. i= For ay -scalars x,, x we have xα+ K x α = x α j j j= = = = m x A β j ij i j= i= m j= i= (A x ) β ij j i β m Aij xj. i i= j= Sice > m we see here exiss scalars x,, x o all zero such ha Ax ij j = 0; i m j= Hece x α + + x α = 0 which proves S is a liearly depede se. The immediae cosequece of his heorem is ha ay wo basis of a fiie dimesioal super vecor space have same umber of elemes. As i case of usual vecor space whe we say a supervecor space is fiie dimesioal i has fiie umber of elemes i is basis. We illusrae his siuaio by a simple example. Example.2.28: Le V = {(x x 2 x 3 x 4 ) x i Q; i 4} be a super vecor space over Q. I is very clear ha V is fiie 50

52 dimesioal ad has oly four elemes i is basis. Cosider a se S = {( 0 0), ( 2 3 4), ( ), (0 2 ) ad ( 2 0 2)} {x, x 2, x 3, x 4, x 5 } V, o S is a liearly depede subse of V; i.e. o show his we ca fid scalars c, c 2, c 3, c 4 ad c 5 i Q o all zero such ha cx 0 i i =. c ( 0 0) + c 2 ( 2 3 4) + c 3 ( ) + c 4 ( 0 2 ) + c 5 ( 2 0 2) = ( ) gives c + c 2 + 4c 3 + c 5 = 0 2c 2 + c 4 + 2c 5 = 0 c + 3c 2 + 2c 4 = 0 4c 2 + 3c 3 + c 4 + 2c 5 = 0. I is easily verified we have o zero values for c,, c 5 hece he se of 5 super vecors forms a liearly depede se. I is lef as a exercise for he reader o prove he followig simple lemma. LEMMA.2.: Le S be a liearly idepede subse of a super vecor space V. Suppose β is a vecor i V ad o i he super subspace spaed by S, he he se obaied by adjoiig β o S is liearly idepede. We sae he followig ieresig heorem. THEOREM.2.6: If W is a super subspace of a fiie dimesioal super vecor space V, every liearly idepede subse of W is fiie ad is par of a (fiie basis for W). Sice super vecors are also vecors ad hey would be coribuig more elemes while doig furher operaios. The above heorem ca be give a proof aalogous o usual vecor spaces. 5

53 Suppose S 0 is a liearly idepede subse of W. If S is a liearly idepede subse of W coaiig S 0 he S is also a liearly idepede subse of V; sice V is fiie dimesioal, S coais o more ha dim V elemes. We exed S 0 o a basis for W as follows: S 0 spas W, he S 0 is a basis for W ad we are doe. If S 0 does o spa W we use he precedig lemma o fid a super vecor β i W such ha he se S = S 0 { β } is idepede. If S spas W, fie. If o, we apply he lemma o obai a super vecor β 2 i W such ha S2 = S { β 2} is idepede. If we coiue i his way he (i o more ha dim V seps) we reach a a se Sm = S 0 { β, K, βm} which is a basis for W. The followig wo corollaries are direc ad is lef as a exercise for he reader. COROLLARY.2.: If W is a proper super subspace of a fiie dimesioal super vecor space V, he W is fiie dimesioal ad dim W < dim V. COROLLARY.2.2: I a fiie dimesioal super vecor space V every o empy liearly idepede se of super vecors is par of a basis. However he followig heorem is simple ad is lef for he reader o prove. THEOREM.2.7: If W ad W 2 are fiie dimesioal super subspaces of a super vecor space V he W + W 2 is fiie dimesioal ad dim W + dim W 2 = dim (W W 2 ) + dim (W + W 2 ). We have see i case of super vecor spaces we ca defie he elemes of hem as m super marices or as super row vecors or as super colum vecors. 52

54 So how o defie liear rasformaios of super vecor spaces. Ca we have liear rasformaios from a super vecor space o a super vecor space whe boh are defied over he same field F?.3 Liear Trasformaio of Super Vecor Spaces For us o have a meaigful liear rasformaio, if V is a super vecor space, super row vecors havig compoes (A,, A ) where each A i a is row vecor of same legh he we should have W also o be a super vecor space wih super row vecors havig oly compoes of some legh, eed o be of ideical legh. Whe we say wo super vecor have same compoes we mea ha boh he row vecor mus have same umber of pariios. For isace X = (x, x 2,, x ) ad Y = (y, y 2,, y m ), m, he umber of pariios i boh of hem mus be he same if X = (A A ) he Y = (B B ) where A i s ad B j s are row vecors i, j. Le X = ( ) ad Y = ( ). If X is pariioed as X = ( ) ad Y = ( ). X = (A A 2 A 3 ) ad Y = (B B 2 B 3 ) where A = 2, B = ( 0 2); A 2 ( 0 5), B 2 = (3 4 5), A 3 = (6 ) ad B 3 = (7 8 ). We say he row vecors X ad Y have same umber of pariios or o be more precise we say he super vecors have same umber of pariios. We ca defie liear rasformaio bewee wo super vecor spaces. Super vecors wih same umber of elemes or wih same umber of pariio of he row vecors; oherwise we cao defie liear rasformaio. Le V be a super vecors space over he field F wih super vecor X V he X = (A A ) where each A i is a row vecor. Suppose W is a super vecor space over he same field F 53

55 if for a super row vecor, Y W ad if Y = (B B ) he we say V ad W are super vecor spaces wih same ype of super row vecors or he umber of pariios of he row vecors i boh V ad W are equal or he same. We call such super vecor spaces as same ype of super vecor spaces. DEFINITION.3.: Le V ad W be super vecor spaces of he same ype over he same field F. A liear rasformaio from V io W is a fucio T from V io W such ha T(cα + β) = ctα + β for all scalars c i F ad he super vecors α, β V; i.e. if α = (A A ) V he Tα = (B B ) W, i.e. T acs o A i such a way ha i is mapped o B i.e. firs row vecor of α i.e. A is mapped io he firs row vecor B of Tα. This is rue for A 2 ad so o. Uless his is maiaied he map T will o be a liear rasformaio preservig he umber of pariios. We firs illusrae i by a example. As our mai aim of iroducig ay oio is o for givig ice defiiio bu our aim is o make he reader udersad i by simple examples as he very cocep of super vecors happe o be lile absrac bu very useful i pracical problems. Example.3.: Le V ad W be wo super vecor spaces of same ype defied over he field Q. Le V = {(x x 2 x 3 x 4 x 5 x 6 ) x i Q; i 6} ad W = {(x x 2 x 3 x 4 x 5 x 6 x 7 x 8 ) x i Q; i 8}. We see boh of hem have same umber of pariios ad we do o demad he legh of he vecors i V ad W o be he same bu we demad oly he legh of he super vecors o be he same, for here we see i boh he super vecor spaces V ad W super vecors are of legh 3 oly bu as vecors V has aural legh 6 ad W has aural legh 8. Le T : V W 54

56 T (x x 2 x 3 x 4 x 5 x 6 ) = (x + x 3 x 2 + x 3 x 4 x 4 + x 5 x 5 x 6 0 x 6 ). I is easily verified ha T is a liear rasformaio from V io W. Example.3.2: Suppose V = {(x x 2 x 3 x 4 x 5 ) x,, x 5 Q} ad W = {(x x 2 x 3 x 4 x 5 ) x i Q; i 5} boh super vecor spaces over F. Suppose we defie a map T : V W by T[(x x 2 x 3 x 4 x 5 )] = (x + x 2 x 3 + x 4, x 5 0 0). T is a liear rasformaio bu does o preserve pariios. So such liear rasformaio also exiss o super vecor spaces. Example.3.3: Le V = {(x x 2 x 3 x 4 x 5 x 6 ) x i Q; i 6} ad W = {(x x 2 x 3 x 4 ) x i Q, i 4}. The we cao defie a liear rasformaio of he super vecor spaces V ad W. So we demad if we wa o defie a liear rasformaio which is o pariio preservig he we demad he umber pariio i he rage space (i.e. he super vecor space which is he rage of T) mus be greaer ha he umber of pariios i he domai space. Thus wih his demad i mid we defie he followig liear rasformaio of wo super vecor spaces. DEFINITION.3.2: Le V = {(A A 2 A ) A i row vecors wih eries from a field F} be a super vecor space over F. Suppose W = {(B B 2 B m ), B i row vecors from he same field F; i =, 2,, m} be a super vecor space over F. Clearly m. The we call T he liear rasformaio i.e. T: V W where T(A i ) = B j, i ad j m ad eries B K i W which do o have a associaed A i i V are jus pu as zero row vecors ad if T is a liear rasformaio from A i o B j ; T is called as he liear rasformaio which does o preserve pariio bu T acs more like a embeddig. Oly whe m = we ca defie he oio of pariio preservig liear rasformaio of super vecor spaces from V io W. Bu whe 55

57 > m we will o be i a posiio o defie liear rasformaio from super vecor space V io W. Wih hese codiios we will give ye some more examples of liear rasformaio from a super vecor space V io a super vecor space W boh defied over he same field F. Example.3.4: Le V = {(x x 2 x 3 x 4 x 5 x 6 x 7 ) x i Q; < i < 7} be a super vecor space over Q. W = {(x x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 ) x i Q; < i < 9} be a super vecor space over Q. Defie T : V W by T (x x 2 x 3 x 4 x 5 x 6 x 7 ) (x + x 2 x 2 + x 3 x 3 + x 4 x 5 + x 6 x x 9 ). I is easily verified T is a liear rasformaio from V o W, we ca have more umber of liear rasformaios from V o W. Clearly T does o preserve he pariios. We also oe ha umber of pariios i V is less ha he umber of pariios i W. We give ye aoher example. Example.3.5: Le T : V W be a liear rasformaio from V io W; where V = {(x x 2 x 3 x 4 x 5 x 6 ) x i Q; i 6} is a super vecor space over Q. Le W = {(x x 2 x 3 x 4 x 5 x 6 x 7 ) x i Q; i 7} be a super vecor space over Q. Defie T ((x x 2 x 3 x 4 x 5 x 6 ) = (x + x 2 x 2 + x 3 x 2 x 4 + x 5 x 5 + x 6 x 6 + x 4 x 4 + x 5 + x 6 ) I is easily verified ha T is a liear rasformaio from he super vecor space V io he super vecor space W. Now we proceed io defie he kerel of T or ull space of T. DEFINITION.3.3: Le V ad W be wo super vecor spaces defied over he same field F. Le T : V W be a liear rasformaios from V io W. The ull space of T which is a super subspace of V is he se of all super vecors α i V such ha Tα = 0. I is easily verified ha ull space of T; N = {α V T(α) = 0} is a super subspace of V. For we kow T(0) = 0 so N is o empy. 56

58 If he Tα = Tα 2 = 0 T(cα + α 2 ) = ctα + Tα 2 = c = 0. So ha for every α α 2 N, cα + α 2 N. Hece he claim. We see whe V is a fiie dimesioal super vecor space he we see some ieresig properies relaig he dimesio ca be made as i case of vecor spaces. Now we proceed o o defie he oio of super ull subspace ad he super rak space of a liear rasformaio from a super vecor space V io a super vecor space W. DEFINITION.3.4: Le V ad W be wo super vecor spaces over he field F ad le T be a liear rasformaio from V io W. The super ull space or ull super space of T is he se of all super vecors α i V such ha Tα = 0. If V is fiie dimesioal, he super rak of T is he dimesio of he rage of T ad ulliy of T is he dimesio of he ull space of T. This is rue for boh liear rasformaios preservig he pariio as well as he liear rasformaios which does o preserve he pariio. Example.3.6: Le V = {(x x 2 x 3 x 4 x 5 ) x i Q; i 5} be a super vecor space over Q ad W = {(x x 2 x 3 x 4 ) x i Q; i 4} be a super vecor space over Q. Le T: V W defied by T(x x 2 x 3 x 4 x 5 ) = (x + x 2, x 2 x 3 + x 4, x 4 + x 5 ). T is easily verified o be a liear rasformaio. The ull super subspace of T is N = {(0 0 k, k, k) k Q} which is a super subspace of V. Now dim V = 5 ad dim W = 4. Fid dim N ad prove rak T + ulliy T = 5. Suppose V is a fiie dimesioal super vecor space over a field F. We call B = {x,, x } o be a basis of V if each of he x i s are super vecors from V ad hey form a liearly idepede se ad spa V. Suppose V = {(x 57

59 x ) x i Q; i } he dimesio of V is ad V has B o be is basis he B has oly -liearly idepede elemes i i which are super vecors. So i case of super vecor spaces he basis B forms a se which coais oly supervecors. THEOREM.3.: Le V be a fiie dimesioal super vecor space over he field F i.e., V = {(x x 2 x ) x i F; i } Le (α, α 2,, α ) be a basis of V i.e., each α i is a super vecor; i =, 2,,. Le W = {(x x m ) x i F; i m} be a super vecor space over he same field F ad le β,, β be super vecors i W. The here is precisely oe liear rasformaio T from V io W such ha Tα j = β j ; j =, 2,,. Proof: To prove ha here exiss some liear rasformaio T from V io W wih Tα j = β j we proceed as follows: Give α i V, a super vecor here is a uique -uple of scalars i F such ha α = x α + + x α where each α i is a super vecor ad {α,, α } is a basis of V; ( i ). For his α we defie Tα = x β + + x β. The T is well defied rule for associaig wih each super vecor α i V a super vecor Tα i W. From he defiiio i is clear ha Tα j = β j for each j. To show T is liear le β = y α + + y α be i V for ay scalar c F. We have cα + β = (cx + y ) α + + (cx + y ) α ad so by defiiio T (cα +β) = (cx + y ) β + + (cx + y ) β. O he oher had c(tα) + Tβ = c xiβi + yiβi = i= i= i= (cx + y ) β i i i ad hus T (cα + β) = c(tα) + Tβ. If U is a liear rasformaio from V io W wih Uα j = β j ; j =, 2,, he for he super vecor α = xiαi we have Uα = i= 58

60 U ( xiαi ) = i= xu i αi = i= x iβi, so ha U is exacly he rule i= T which we have jus defied above. This proves ha he liear rasformaio wih Tα j = β j is uique. Now we prove a heorem relaig rak ad ulliy. THEOREM.3.2: Le V ad W be super vecor spaces over he field F of same ype ad le T be a liear rasformaio from V io W. Suppose ha V is fiie-dimesioal. The super rak T + super ulliy T = dim V. Proof: Le V ad W be super vecor spaces of he same ype over he field F ad le T be a liear rasformaio from V io W. Suppose he super vecor space V is fiie dimesioal wih {α,, α k } a basis for he super subspace which is he ull super space N of V uder he liear rasformaio T. There are super vecors {α k+,, α } i V such ha {α,, α } is a basis for V. We shall prove {Tα k+,, Tα } is a basis for he rage of T. The super vecors {Tα k+,, Tα } ceraily spa he rage of T ad sice Tα j = 0 for j k, we see Tα k+,, Tα spa he rage. To prove ha hese super vecors are liearly idepede; suppose we have scalars c i such ha c(t i αi) = 0. This says ha vecor α = i= k+ T( c iαi) = 0 ad accordigly he super i= k+ ciαi is i he ull super space of T. Sice α,, i= k+ α k form a basis of he ull super space N here mus be scalars b,, b k such ha α = k b α. Thus i i i= k biαi i= c jαi = 0 j= k+ 59

61 Sice α,, α are liearly idepede we mus have b = b 2 = = b k = c k+ = = c = 0. If r is he rak of T, he fac ha Tα k+,, Tα form a basis for he rage of T ells us ha r = k. Sice k is he ulliy of T ad is he dimesio of V, we have he required resul. Now we wa o disiguish he liear rasformaio T of usual vecor spaces from he liear rasformaio T of he super vecor spaces. To his ed we shall from here owards deoe by T s he liear rasformaio of a super vecor space V io a super vecor space W. Furher if V = {(A A ) A i are row vecors wih eries from F, a field} ad V a super vecor space over he field F ad W = {(B B ) B i are row vecors wih eries from he same field F} ad W is also a super vecor space over F. We say T s is a liear rasformaio of a super vecor space V io W if T = (T T ) where T i is a liear rasformaio from A i o B i ; i =, 2,,. Sice A i is a row vecor ad B i is a row vecor T i (A i ) = B i is a liear rasformaio of he vecor space wih collecio of row vecors A i = (x x i ) wih eries from F io he vecor space of row vecors B i wih eries from F. This is rue for each ad every i; i =, 2,,. Thus a liear rasformaio T s from a super vecor space V io W ca iself be realized as a super liear rasformaio as T s = (T T ). From here o words we shall deoe he liear rasformaio of fiie dimesioal super vecor spaces by T s = (T T 2 T ) whe he liear rasformaio is pariio preservig i case of liear rasformaio which do o preserve pariio will also be deoed oly by T s = (T T 2 T ). Now if (A A ) V he T(A A ) = (T A T 2 A 2 T A ) = (B B 2 B ) W i case T s is a pariio preservig liear rasformaio. If T s is o a pariio preservig rasformaio ad if (B B m ) W we kow m > so T(A A ) = (T A T A ) = (T A 0 0 T 2 A T A ) i whichever maer he liear rasformaio has bee defied. 60

62 THEOREM.3.3: Le V = {(A A ) A i s are row vecors wih eries from F ; i } a super vecor space over F. W = {(B B ) B i s are row vecors wih eries from F; i } a super vecor space over F. Le T s = (T T ) ad U s = (U U ) be liear rasformaios from V io W. The fucio T s + U s = (T + U T + U ) defied by (T s + U s ) (α) = (T s + U s ) (A A ) (where α V is such ha α = (A A ) = (T A + U A T 2 A 2 + U 2 A 2 T A + U A ) is a liear, rasformaio from V io W. If d is ay eleme of F, he fucio dt = (dt dt ) defied by (dt) (α) = d(tα) = d(t A T A ) is a liear rasformaio from V io W, he se of all liear rasformaios from V io W ogeher wih addiio ad scalar muliplicaio defied above is a super vecor space over he field F. Proof: Suppose T s = (T T ) ad U s = (U U ) are liear rasformaios of he super vecor space V io he super vecor space W ad ha we defie (T s + U s ) as above he (T s + U s ) (dα + β) = T s (dα + β) + U s (dα + β) where α = (A A 2 A ) ad β = (C C ) V ad d F. (T s = U s ) (dα + β) = (T + U T + U ) (da + C da 2 + C 2 da + C ) = (T (da + C ) T 2 (da 2 + C 2 ) T (da + C )) + (U (da + C ) U (da + C )) = T (da + C ) + U (da + C ) T (da + C ) + U (da +C )) = (dt A + T C dt A + T C ) + (du A + U C ) du A + U C ) = (dt A dt A ) + (T C T C ) + (du A du A ) + (U C U C ) = (dt A dt A ) + (du A du A ) + (T C T C ) + (U C U C ) = (d(t +U )A d(t + U ) A ) + ((T +U ) C (T + U ) C ) 6